The solubility product constant of silver bromide is 5.0 × 10^-13.What is the molar solubility of this compound?

Question

Quizplus · Accepted Answer

The molar solubility (x) of silver bromide (AgBr) can be calculated using the solubility product constant (Ksp) expression:

Ksp = [Ag+][Br-] = 5.0 × 10^-13

Since AgBr is a 1:1 electrolyte,

[Ag+] = x M 
[Br-] = x M

Substituting these values in the Ksp expression:

Ksp = x^2

x = sqrt(Ksp) = sqrt(5.0 × 10^-13)

x = (50)^(1/2) × 10^-7 M

x = (50)^(1/2) × 0.0000001 M

x = (50)^(1/2) × 10^-6 M

x ≈ 7.1 × 10^-7 M

Therefore, the molar solubility of AgBr is approximately 7.1 × 10^-7 M. This corresponds to choice C.

The Solubility Product Constant of Silver Bromide Is 5